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December 7

Radius of intersection of sphere and plane

Hello. THe problem is "What is the radius of the circle that is the intersection of the plane $x+y+z=1$ and the sphere $x^{2}+y^{2}+z^{2}=1$ ?" I set the equations equal, then set about completing the square so that I got $(x-{\frac {1}{2}})^{2}+(y-{\frac {1}{2}})^{2}+(z-{\frac {1}{2}})^{2}={\frac {3}{4}}$ . This gave me a sphere of radius ${\frac {\sqrt {3}}{2}}$ but this wasn't one of the answers. I'm not sure how to approach this problem. Help? 24.92.74.238 (talk) 02:36, 7 December 2012 (UTC)[reply]

Quick and dirty... The points (1,0,0), (0,1,0) and (0,0,1) are all on the edge of the circle (no intersection point can have a value for any co-ordinate greater than one). This is an equilateral triangle inscribed in the circle that you want (which is the same as saying the circle you want circumscribes that triangle). The distance between each of those points is

{\sqrt {2}}

. According to Equilateral Triangle, if the edges of the equilateral triangle are of length a, the radius of the circumscribed circle is a *

{\frac {\sqrt {3}}{3}}

. So the result is

{\frac {\sqrt {6}}{3}}

=

{\frac {\sqrt {2}}{\sqrt {3}}}

Are either of these the answer?Naraht (talk) 05:11, 7 December 2012 (UTC)[reply]

You should take care of what your equations mean. The first means "all points (x,y,z) for which x + y + z = 1", and the second means "all points (x,y,z) for which x² + y² + z² = 1". You are asked to find the intersection of these two sets, i.e. "all points (x,y,z) for which x + y + z = 1 and simultaneously x² + y² + z² = 1". From geometric considerations you will know that the intersection of a sphere and a plane is a circle (or a point or the empty set, depending on the relative separation). So arriving at an equation for a sphere rather than a circle, already you should know that you have discarded a constraint, and ended up with a sphere that happens to contain the circle of intersection, but you've lost the information of which circle.

You could try various approaches. This problem is simple to deal with geometrically due to the particular choice of coefficients. Visualize the actual plane and the actual sphere relative to the axes. You could rotate these (using a substitution of variables) so that the plane is parallel to two of the axes, or else work out some obvious points where these figures intersect (values of x, y and z that simultaneously satisfy both equations), and work geometrically from there. — Quondum 05:22, 7 December 2012 (UTC)[reply]

I corrected the title from "point" to "sphere". StuRat (talk) 05:37, 7 December 2012 (UTC) [reply]

I think my quick and dirty method above works, but only because there are nice integer points in the intersection. If the sphere had had a radius of 1.5 instead, my method is useless....Naraht (talk) 18:29, 7 December 2012 (UTC)[reply]

Not necessarily. How about if you solve the linear constraint for z in terms of x and y and substitute in for z in the quadratic constraint. Then pick 3 (x,y) pairs that satisfy the resulting quadratic in x and y and plug each of those back into the solution for z. This should give three points on the desired circle, and we can use your very elegant method (using the formula for circumradius of a triangle with general sides). (I haven't lost any information handling the constraints this way, have I?) Duoduoduo (talk) 20:08, 7 December 2012 (UTC)[reply]

Well maybe not useless since we have the general triangle circumradious, but no where near quick and dirty. to use my changed example, that means picking 3 x,y pairs that fit 1.5 = x^2 + y^2 + (1-x-y)^2 = x^2 + y^2 + (1+x^2+y^2-2x-2y-2xy)= 1+2x^2+2y^2-2x-2y-2xy. I can't figure out any x,y that fits that off the top of my head, good luck...Naraht (talk) 20:26, 7 December 2012 (UTC)[reply]

Maybe not quick and dirty, but an alternative to Quondom's suggestion to rotate by changing variables. Duoduoduo (talk) 21:49, 7 December 2012 (UTC)[reply]

Given the problem as a homework or exam question, I would have used Naraht's "quick and dirty" method without a doubt, which is what I had in mind when I said "else work out some obvious points ... and and work geometrically from there"; I was trying to encourage some thinking (and hadn't noticed that Naraht had added an answer before I'd opened my edit window). The more general methods (the rotation or even finding a parameterized curve) are rather laborious, but it may help to be aware of them. — Quondum 09:40, 8 December 2012 (UTC)[reply]

You could find the point on plane closest to origin. A then find the distance, d, from that point to the origin. Then you have a right angled triangle with the radius of the sphere, 1, as the hypothenuse and d and r the radius of the circle as the other two sides. This can be solved by Pythagoras. In this example the point is $(1/3,1/3,1/3)$ and $d=1/sqrt(3)$ . The radius of the circle is then $r=sqrt(1^{2}-1/3)=sqrt(2/3)$ . This approach should be easy to generalise.--Salix (talk): 10:37, 8 December 2012 (UTC)[reply]

Mazes (complex branching passages)

Do mathematicians have a method for measuring and comparing the complexities of mazes (complex branching passages)? These are multicursal passages, as distinct from labyrinths (unicursal passages).
—Wavelength (talk) 03:34, 7 December 2012 (UTC)[reply]

Don't know if there is one already, but we can invent one. How about if we add up the number of 3-way branches and 4-way branches (maybe count those doubly) ? Of course, if the entrance and exit are right next to each other, with no wall in-between, then a seemingly complex maze can be solved easily. Perhaps we should just count the minimum number of 3-way and 4-way branches we must encounter to get from the starting point to the end. StuRat (talk) 01:48, 8 December 2012 (UTC)[reply]

(Note #1) My preference is for a method that is already used by mathematicians, although I myself can probably devise a variety of methods involving various properties of mazes. (Note #2) My question did not specify mazes of the type found in popular puzzles, where each maze has exactly one entrance and exactly one exit.

Different types of branches (triple, quadruple, and quintuple) might increase complexity, but interior island configurations might be analyzed as decreasing complexity inasmuch as they increase the number of practical routes.

—Wavelength (talk) 16:40, 8 December 2012 (UTC)[reply]

It would be quite hard to assign a "natural" (i.e. not human-centric) complexity to a maze, seeing as any maze can be solved in linear time and space with a simple graph search using any arbitrary rule for pulling from the fringe (e.g. DFS, BFS). That is, the only way to make a maze that's really fundamentally hard is to make it bigger.

On the other hand, if you're willing to accept non-natural algorithms for assigning difficulty values to mazes, here are some ideas (for mazes with a single start point and single end point):

Run a depth-first search, using a distance heuristic with bounded lookahead to choose the next visited vertex from each point. Count the number of steps of backtracking the algorithm does before it finds a solution.
Run an A* search with a distance heuristic. Divide the number of visited vertices by the minimum path length.

Both of these have desirable properties like giving small numbers on labyrinths with no branching, reducing complexity when there are islands permitting multiple valid paths to the endpoint, etc. « Aaron Rotenberg « Talk « 07:51, 10 December 2012 (UTC)[reply]

Thank you for your reply.

—Wavelength (talk) 20:22, 10 December 2012 (UTC)[reply]

Does graph theory have such a method, so that, for example, one maze might have a complexity of 2 and another maze might have a complexity of 3?
—Wavelength (talk) 18:21, 7 December 2012 (UTC)[reply]

Possibly the most important measure of complexity is whether the maze is a simple tree which has branches and no loops or one with cycles.--Salix (talk): 10:17, 8 December 2012 (UTC)[reply]

I'm with Salix on this one. I'd convert mazes into graphs first, by some trivial method, and then run optimizations on that.

.Maze                Graph
.#############
.#   # #   #F#    +-+ + +-+ F
.# # # ### # #    | | |   | |
.# #S#     # #    + S +-+-+ +
.# ### # ### #    |   | |   |
.#     # #   #    +-+-+ + +-+
.# # ##### # #    | |     | |
.# #         #    + +-----+-+
.#############

I did use only one start and one finish square here, but if you don't like that, you could label any square the way you want. In further processing, thy are just two labeled nodes, S and F respectively. If you remove all unlabeled nodes with exactly 2 neighbors, you get the following graph:

Before          After
.+-+ + +-+ F        + +   F
.| | |   | |        | |   |
.+ S +-+-+ +      S +-+   |
.|   | |   |     / /  |   |
.+-+-+ + +-+    +-+   + +-+
.| |     | |    |  \____| |
.+ +-----+-+    +       +-+

The 2x2 nodes close to F are special; if we removed two of them, we would end up with two edges, both of which would connect the same two points. We can, however, remove one of them:

.    + +   F
.    | |   |
.  S +-+   |
. / /  |   |
.+-+   +   +
.|  \____ /|
.+       +-+

The number of edges could then be seen as a measure of the complexity. The number of surviving edges after all unlabeled nodes with only one or two neighbors have been eliminated would be another measure of complexity. Only S, F, the three nodes of the triangle, and five edges would survive. If the maze is a deformed tree, only S and F and the connecting edge will survive.

Not sure if I've added anything you don't already know, though. - ¡Ouch! (^{hurt me} / _{more pain}) 08:05, 11 December 2012 (UTC)[reply]

Thank you for your reply with diagrams. I should have searched for an answer online earlier, but I only began to do so today, and from a search for measure maze complexity, I found a method explained at http://www.math.uco.edu/mcclendon/complexityrecmazes.pdf. The calculations there involve summations and fractions. Also, I have discovered the field known as computational complexity theory, so this discussion involves the intersection of graph theory and computational complexity theory.

—Wavelength (talk) 18:03, 11 December 2012 (UTC)[reply]

December 8

X chromosomes and statistics

This article (PDF) makes the claims that 1) variance in intelligence among men is higher, and 2) this is due to the fact that men have one X chromosome (which has a disproportionately large number of genes expressed in the brain) and women have two, which get "averaged" during genetic cross-over.

I can believe 1), but I'm having trouble accepting 2). If every man gets his X chromosome from his mother, then would the variance in intelligence among men be the same as that of women? 65.92.7.202 (talk) 01:16, 8 December 2012 (UTC)[reply]

Different genes working in different ways. For some genes, one is dominant, and the recessive gene is not expressed at all. Eye color tends to work this way. For others, you get an averaging of the two genes, which seems to be how they claim intelligence works. So, if the woman has two X chromosomes, one of which would give her an IQ of 120, and one of which would give her an IQ of 80, then she ends up with an IQ of 100 (gross oversimplification, of course). However, if she only passes one gene on to her son, he will either have the IQ of 80 or 120. StuRat (talk) 01:32, 8 December 2012 (UTC)[reply]

I've fixed the OP's link. I agree fully with the explanation. One does of course have to be aware of the uncertainty in the assumptions on the averaging effect with duplicate genes, and also that there will be contributing factors of an unrelated origin; however, given the assumption, this mechanism can plausibly be invoked as a possible contributor to greater variability of some characteristics in men. — Quondum 10:15, 8 December 2012 (UTC)[reply]

Also note that if we use their assumptions, then fathers must not contribute to the intelligence of their sons, as they don't provide an X chromosome to them. I'd think a simple study would show that this is not true. You'd need to study sperm donors, though, to ensure that the males are only providing genetic contributions, as opposed to affecting the environment in which their son is raised. StuRat (talk) 16:45, 8 December 2012 (UTC)[reply]

But what about X-inactivation? 65.92.7.202 (talk) 23:35, 8 December 2012 (UTC)[reply]

No, their assumptions would only imply that fathers have a smaller genetic contribution to their sons than do mothers.--80.109.106.49 (talk) 21:57, 8 December 2012 (UTC)[reply]

Since the OP is aware of X-inactivation, the OP is undoubtedly aware that only one of the two X chromosomes in women is expressed -- the other is inactivated. This makes the claim pretty implausible, unless I am missing something. Looie496 (talk) 00:49, 9 December 2012 (UTC)[reply]

Wonderful link, thanks for it. But I don't understand your interpretation of it in this context. My reading of the article is that in any given cell only one of the X chromosomes is expressed, but that each of the X chromosomes is expressed in some cells. To me it would seem then that averaging could occur if intelligence depends on what's happening in many cells. Duoduoduo (talk) 12:39, 9 December 2012 (UTC)[reply]

Math of the McEliece cryptosystem

It looked simple enough on this page, but I found out that the matrixes are not normal ones. Can someone explain to me (a person with no knowledge of how a Hamming code works) help me to understand how to do that page's example out by hand? --Melab±1 ☎ 07:28, 8 December 2012 (UTC)[reply]

You seem to be asking for a tutorial, which may be more than most would be prepared to do. You have also not shown where you got stuck. You'll have to get a feel for error correction codes before you tackle this. Your comment "the matrixes are not normal ones" might suggest that you have not realized that all arithmetic is done modulo 2. — Quondum 10:29, 8 December 2012 (UTC)[reply]

Then if I were to multiply two matrices I would need to apply modulo 2 to every step in calculating the values for each position in the matrix? --Melab±1 ☎ 00:35, 9 December 2012 (UTC)[reply]

Yep, or you can also just do standard integer multiplication in the real numbers and then reduce the answer modulo 2 at the very end. Reducing at every step will keep the numbers smaller throughout the multiplication. - Looking for Wisdom and Insight! (talk) 06:37, 10 December 2012 (UTC)[reply]

Or alternately regard every element as a boolean value. Addition then is replaced with exclusive or, and multiplication with logical and. Inverting a matrix is quite fun – Gaussian elimination still works, but the only value you ever have to divide by is 1. — Quondum 08:05, 10 December 2012 (UTC)[reply]

Pump capacity formulas

HI, I'M LOOKING FOR MATH FORMULAS CONCERNING RECIPROCATING PUMP CAPACITY (FOR U.S. COAST GUARD MERCHANT MARINE UPGRADE TEST). A SIMPLEX SINGLE/DOUBLE ACTING RECIPROCATING PUMP MAKING 180 STROKES/MINUTE, WITH A 5" DIAMETER CYLINDER, AN 8" STROKE AND OPERATING WITH 87% VOLUMETRIC EFFICIENCY, WHAT IS THE CAPACITY OF THE PUMP? ESB — Preceding unsigned comment added by Edgarbarrios (talk • contribs) 15:43, 8 December 2012 (UTC)[reply]

The volume of a cylinder is:

V = hπr²

They gave you a diameter of 5 inches, so the radius is half that, or 2.5 inches. Presumably you want the figures in cubic feet, so we convert that to 2.5/12 (or 5/24) foot. Similarly, the stroke is the same as the height, so 8 inches is 2/3 foot. We could just use 3.14 as a reasonable approximation for pi. Do the substitutions:

V = (2/3)(3.14)(5/24)²

You will then need to multiply by the volumetric efficiency, and the strokes per minute:

180(0.87)(2/3)(3.14)(5/24)²

The answer should be in cubic feet per minute. I assume there is only one cylinder. If "double acting" means that both sides are pumping at once, then multiply the above number by 2. Do the math and show us your results, and we will check your work. StuRat (talk) 16:26, 8 December 2012 (UTC)[reply]

December 9

New musical tuning system calculation, math question.

Sound is frequency. In the way we usually tune instruments we start with an frequency of 440hz. Thats A4. A5 is A4 * 2 = 880hz, A6 is A5 *2 = 1760hz...... Between A4 and A5, we have B4, C4..., but this is not needed to my question, so I will not explain it.

The question.

I, was thinking about how, some "logical" tuning system would work. And need your help.

This logical tuning would start with 20 hz (lowest frequency we can hear), and would continue increasing following some formula/calculation/whateaver, until it goes to 440hz. Then it would continue to increase, using THE SAME formula/calculation/whateaver as before until it reach 20000hz (highest frequency we can hear). The amount of steps this tuning needs to go from 20 to 440hz, NEED to be the same amount of steps it would take to go from 440 to 20000

Anyway how this formula could be?. 201.78.126.14 (talk) 02:11, 9 December 2012 (UTC)[reply]

Well, you basically have 3 points on a graph, say (0,20), (50,440), and (100,20000). You can run a circle through any 3 non-collinear points like this. StuRat (talk) 04:11, 9 December 2012 (UTC)[reply]

... or you can fit a quadratic function. Or almost any other formula that involves 3 independent parameters. And you have even more choice if you don't fix the number of steps at 100. Of course there are physical reasons why tuning involves doubling of frequencies, so replacing this with some other convention chosen at random will be less logical, not more logical. Gandalf61 (talk) 05:00, 9 December 2012 (UTC)[reply]

Agreed. If you tried to tune your instrument like that, it would be all forked up. StuRat (talk) 05:03, 9 December 2012 (UTC) [reply]

Yeah, who cares about logic, it would produce hideous dissonance. Looie496 (talk) 16:36, 9 December 2012 (UTC)[reply]

Although other scales are possible, see scale (music). StuRat (talk) 16:40, 9 December 2012 (UTC)[reply]

Why should the number of steps on each side of 440Hz be the same? Why is that frequency special in your tuning system? Before defining a "logical" system, you should look into why instruments are tuned the way they are, and explain what is illogical about them and why. Then you can start working on designing something that solves the problems you've identified. 209.131.76.183 (talk) 19:30, 10 December 2012 (UTC)[reply]

Indeed, 440Hz is a pretty random choice, as can be seen from the history of concert pitch. And the 20Hz and 20000Hz endpoints are fairly random as well, the actual range of hearing varies substantially between people, these are just ballpark figures neatly expressible with the quite randomly chosen unit of frequency and semi-randomly chosen numerical notation system we happen to be using. Trying to base anything “logical” on these three points is rather ridiculous.—Emil J. 20:40, 10 December 2012 (UTC)[reply]

OP HERE: Because 440hz would be the middle point, and so you would have to have the same amount of steps to get from the starting point (when humans start to listen to sounds) to 440hz as you would need to go from 440hz to the endpoint(when humans stop being able to listen to it). About my stuff being logical or not, the idea is that, if 440 is the middle point (or that would be middle C 261.626 Hz?) the amount of steps between the starting point (when we start to listen) and 440 need to be the same as the amount of steps between the middle point and the end point (when we stop listening to sounds). .201.78.166.79 (talk) 22:27, 11 December 2012 (UTC)[reply]

Well I was taught that "middle C" was 256 Hz but either choice cannot be the "middle" of a geometric progression ranging from 20Hz up to 20 kHz. The musical "middle" would need to be the geometric mean of the endpoints. You could make an equal-tempered scale starting at 20 Hz with a middle of 440 Hz if you take your top note to be 9680 Hz (no-one plays notes higher than that, though some overtones go higher). If you require n steps in each half, then each interval ratio would be the nth root of 22. If you try n = 53 or 54 you will get intervals that are approximate semitones. For quarter tones (perhaps the smallest interval clearly heard by most people), try n = 107. By selecting notes from the n = 107 scale, you could get very close to Arab, Indian and Western music (the difference would be noticed only by a highly-trained ear).

Alternatively, for exact semitones, take your lowest note to be 23.542 Hz and your highest note to be 8223.6 Hz and n = 50 (or n = 100 for quarter tones). You might be interested in the article: Microtonal music, but I think all such scales are based on the octave (doubling the frequency), and your exact original specification would not produce a musical scale as we understand music because tunes would sound different depending on the point on the scale at which you started, and you could never play an exact octave. This doesn't mean that you couldn't play "music" of some kind on your musically uneven scale, but it would sound very strange and dissonant unless you selected just certain notes that roughly corresponded to conventional intervals. Dbfirs 18:38, 12 December 2012 (UTC)[reply]

WIkipedia says middle C is 261.626.

You said: "all such scales are based on the octave (doubling the frequency), and your exact original specification would not produce a musical scale as we understand music because tunes would sound different depending on the point on the scale at which you started, and you could never play an exact octave.", yes I know that and I dont care if it the end result is not based on doubling the frequency. By starting with an frequency of 20hz and doubling it, you would go to 40hz, then 80hz, 160, 320, 640. So it would be impossible to go to middle c or 440hz by just doubling the frequency.Also, YES, it needs to "start" with 20hz and "ends" with 20000hz.Also the number of steps not being divisible by 12 is not an problem.201.78.197.147 (talk) 22:47, 12 December 2012 (UTC)[reply]

I wasn't doubting Wikipedia, just reminding you that "middle" was different in the past and is just an arbitrary convention. It is impossible to make a "musical" scale (based on consistent frequency ratios) starting at 20 Hz with 440 Hz in the "middle" and ending at 20 000 Hz, but you could construct an infinite number of different "mathematical" scales with this property (though the simple quadratic and cubic fits go into negative frequency, the circle fit doesn't give a single-valued function, and my attempt at a simple exponential fit doesn't give good results). I suppose the simplest solution is to construct two half-scales with a change in ratios each side of 440 Hz, but you wanted a single formula. An inelegant example of a function that fits your requirement (using 100 intervals, where f is the frequency and n is your note in the scale) is f = 18.0473 plus (4.5334 times n) plus (1.952862 times ten to the power of (n over 25)). You can adjust the constants to get a better fit if you wish, but you will not be able to hear the difference. Some intervals are rather less than a semitone and some are a bit more (but all are less than two semitones except for a few notes below 50Hz). When you create some music using this scale, perhaps you will post your composition so that we can drive away stray cats with it! Perhaps someone more numerate and functionate than I am can come up with a simpler Monotone increasing function through (0,20) (50,440) and (100,20000) with intervals closer to a GP? Dbfirs 09:05, 13 December 2012 (UTC)[reply]

Yes this middle is an arbitrary convention, I knew that before posting the question. The thing about my question/idea is: if this is the middle, the, the number of steps between starting position and ending position should be the same, or it wouldnt be the middle. Also I dont care if is not "musical" (thats subjective anyway). Anyway, I will check the math you posted.189.115.206.17 (talk) 10:30, 14 December 2012 (UTC)[reply]

Tested your formula using wolfram math. With n equal to 50, f=440.004 and not 440, with n=0 f=20.00002 not 20. Anyway thanks, its pretty close to what I wanted.201.78.219.30 (talk) 21:01, 14 December 2012 (UTC)[reply]

Yes, as I said, you can adjust the three constants to get any degree of accuracy you wish, but I stopped there because it is impossible to tune any instrument or analogue sound generator anywhere near this degree of accuracy. Digital sound generators might theoretically generate an exact number of cycles per second, but they cheat, and the result will not be a pure frequency. I'd be interested to hear what your scale sounds like if you compose some music using it. Dbfirs 09:06, 15 December 2012 (UTC)[reply]

"Maybe they're all wrong" ensembles

Do any ensemble learning methods account for correlated errors that can recognize a disagreement between two inner models as evidence against both? An example would be the Bayesian posterior probabilities given that inner classifier X predicted class A, classifier Y independently predicted class B, and both X and Y had the following confusion matrix:

Neon Merlin 04:01, 9 December 2012 (UTC)[reply]

What would be the point of using multiple models if you require them all to agree with each other? Looie496 (talk) 16:30, 9 December 2012 (UTC)[reply]

We don't -- that's the point. In this case, assuming errors are uncorrelated within actual classes (which is probably safe given low noise and heterogeneous inner algorithms), explaining why one predicts A and the other predicts B leads to the conclusion that C is the most probable. Neon Merlin 16:59, 9 December 2012 (UTC)[reply]

Points determining an ellipse or sphere

3 points in the plane (a) are on a circle, (b) which is unique, (c) iff a condition on their relative positions is satisfied: that they not be collinear.

(1) Does there exist n such that n points in the plane (a) are on a conic section, (b) which is unique, (c) iff some condition on their relative positions is satisfied? (Note n>3 since 3 points on an ellipse are also on a circle, making the conic for those 3 points not unique.)

(2) Does there exist k such that k points in R³ (a) are on a sphere, (b) which is unique, (c) iff some condition on their relative positions is satisfied? Duoduoduo (talk) 15:56, 9 December 2012 (UTC)[reply]

Do our five points determine a conic and cubic function articles help ? I don't think R³ can lie on the surface of a sphere, since it's both planar and infinite, and a sphere is neither. StuRat (talk) 16:26, 9 December 2012 (UTC)[reply]

Any 4 points on R³ are on a unique sphere iff there is no plane containing all of them. --77.125.75.226 (talk) 16:52, 9 December 2012 (UTC)[reply]

I see. So those 4 points are on the sphere, while the rest of R³ is not. StuRat (talk) 16:55, 9 December 2012 (UTC)[reply]

(ec)Off the top of my head, your initial statement extends to any number of dimensions (this generalizes the answer to your question (2)): a (k − 2)-sphere is determined by k points. Your condition would be that for k = 4 points, they not be coplanar to determine a (2-)sphere (and the number of Euclidean dimensions does not have to be limited, e.g. 3 points determine a circle in Rⁿ, n ≥ 2). In general, the condition becomes that the k points must not be co-(k − 1)-hyperplanar to determine a (k − 2)-sphere.

A variation is to treat lines, planes etc. as circles, spheres and so on (which they conformally are), which allows the condition to be relaxed. The condition amounts to something like: every point must be separate from the lower-dimensional figure determined by the previous points. E.g., given one point (no constraint), the second point must not be coincident (this determines a 0-sphere, being two points). The third point must be away from the 0-sphere (i.e. not coincident with either of the first two points), and determines a 1-sphere (meaning a circle or line). A fourth point not on the 1-sphere determines a 2-sphere (meaning a sphere or plane). And so on. — Quondum 17:02, 9 December 2012 (UTC)[reply]

Thanks all, for the link and the explanations. Quondum, you say "Your condition would be that for k = 4 points, they not be coplanar to determine a (2-)sphere" and also say "In general, the condition becomes that the k points must not be co-(k − 1)-hyperplanar to determine a (k − 2)-sphere." But it seems to me (maybe I'm missing something) that the latter says that the k=4 points determining a sphere must not be 3-hyperplanar, contrasting with "not be coplanar". Should "not be co-(k − 1)-hyperplanar" be "not be co-(k − 2)-hyperplanar"? Duoduoduo (talk) 17:28, 9 December 2012 (UTC)[reply]

Oops. Yes. (*blush*) — Quondum 17:35, 9 December 2012 (UTC)[reply]

December 10

Impossible math problems

Hi reference desk,

Recently, I came up with a few problems on my own that nobody else could solve. The reason I came up with these problems was partly to test my own progress in learning new concepts in calculus, only to find out much to my dismay that I could not solve them myself. Thus, I'll be posting them here. No, these are not assigned homework problems. In fact, it may turn out that these problems are completely meaningless, have a ridiculously simple and elegant solution or theorem associated, are actually impossible to solve, or all of the above. If the question has no solution or is stupid, please clearly state why. Surely, my conception of these problems is so simple that someone else must have thought of them already, and if such existing articles exist please point me to them. I have thought up similar but easier-to-solve problems in the past, and may post one or two of those in the following weeks, if time permits. For now, I will list only three.

Evaluate $\int _{1}^{e}\ln x-2x^{2}+cos(x^{e})\,dx$ using the Riemann integral sum method.
If $x,y$ are any real finite numbers, find the absolute probability that ${\frac {x}{y}}\,>\,x^{y}$ .
If $g'(x)=f(x)^{-f(x)^{+f(x)^{-f(x)^{+f(x)...}}}}{\text{f(x) times}}$ , compute all possible solutions of $\int f(x)\,dx \over g(x)$ .

Clearly, we may benefit from the following hints:

For a Riemann sum, $R(f,p)={\frac {b-a}{n}}\sum _{i=1}^{n}f(s_{i})$ . From my limited understanding, I turned this into
$R(f,p)={\frac {e-1}{n}}[\sum _{i=1}^{n}ln{\frac {i}{n}}\,-\,\sum _{i=1}^{n}2({\frac {i}{n}})^{2}\,+\,\sum _{i=1}^{n}cos({\frac {i}{n}})^{2}]$ , which simplified to
${\frac {e-1}{n}}\sum _{i=1}^{n}[ln(i)-ln(n)]-{\frac {2e-2}{n^{3}}}\sum _{i=1}^{n}i^{2}+{\frac {e-1}{n}}\sum _{i=1}^{n}cos({\frac {i}{n}})^{2}$ .
This is where I got stuck. Obviously, the Riemann sum method would be very cumbersome to carry out with an integral like this, and I've already made numerous mistakes. For one thing, we would need to know the general formula for the arithmetic sum of $i^{m}$ , where m is any real integer. Note that i refers to an interval, not the complex number ${\sqrt {-1}}$ . For the second sum of our terms, $\sum _{i=1}^{n}i^{2}$ $=$ $n(n+1)(2n+1) \over 6$ , so we can simply to $-{\frac {2e-2}{n^{3}}}\sum _{i=1}^{n}i^{2}={\frac {(1-e)(n+1)(2n+1)}{3n^{2}}}$ . However, the problem underlies what we can do with the arithmetic sum for logarithmic expressions like ln, or even sinusoidal expressions like cos? The only thing that I know is that 0 < ln(x) < 1 when 1 < x < e. Wolframalpha tells me that the integral of this total function is negative over [1, e], so perhaps we should reverse the integral. In either case, we can use the Fundamental theorem of calculus, part II, to show that
$\int _{1}^{e}\ln x-2x^{2}+cos(x^{e})\,dx$ equals, erm, ignoring all constants, and after evaluating the third term not by substitution, but once again by Wolfram...
$xln(x)-x-{\frac {2x^{3}}{3}}-{\frac {x(x^{2e})^{-{\frac {1}{e}}}((-ix^{e})^{\frac {1}{e}}{\Gamma }({\frac {1}{e}},ix^{e})+(ix^{e})^{\frac {1}{e}}{\Gamma }({\frac {1}{e}},-ix^{e}))}{2e}}$ , where Γ(a,x) is the incomplete gamma function. Again, I have no idea what this means.
We know that if we set a closed interval of [x] and [y], we could find the probability that one term is greater than the other in terms of an absolute ratio. However, we soon see that this is no regular diophantine equation. Plotting on wolfram, we get a three-dimensional quasi-supersymmetrical surface! We could similarly select two numbers x and y, for instance x = 1/2 and y = 3, and find that x/y > x^y. When we set $x={\frac {1}{2}},y={\sqrt {2}}$ , we find x/y < x^y. However, this tells us nothing about probability over a given double interval, nor what happens when the intervals overlap, and certainly not enough information about the real number line as a whole, unless we can apply the reversability of 1 and the transcendence of prime-irrationals.
This last one may well be a hoax, and if it is, you may disregard it, but if it is not, then continuity does not imply differentiability. In fact, I think it was a botched attempt at creating a partial differential equation.

Any insight would be greatly appreciated here. Thanks. ~AH1 ^(discuss!) 07:31, 10 December 2012 (UTC)[reply]

2) I'd think the key would be to find the intersections of z = x/y and z = x^y. That should give you some insight. I'd guess that one or the other of those equations is only lower over a finite range of x and y values, with the other being lower for the rest. Thus, the probability will be either 0 or 1. StuRat (talk) 08:03, 10 December 2012 (UTC)[reply]

For number 3, the "f(x) times" part doesn't make much sense for non-integer values of f(x). Your "continuity does not imply differentiability" makes me think you would be interested in reading Weierstrass function. 209.131.76.183 (talk) 19:26, 10 December 2012 (UTC)[reply]

Ratio of x / y > x^y for integers excluding 0

If I plot these ratios for increasing values of x and y I get something approaching 0.372 as x and y approach infinity. Is there a closed form expression for the value of this ratio as x and y approach infinity? 149.169.218.138 (talk) 18:23, 10 December 2012 (UTC)[reply]

What do you mean by x and y approaching infinity? If x is fixed and y goes to infinity, then the lhs tends to 0, and the rhs to infinity. If y is fixed and x goes to infinity, then both lhs and rhs go to infinity, and the ratio lhs/rhs goes to 0 (as long as y > 1). If x and y approach infinity along the line x = ay, where a > 0, then the lhs is constant a, and the rhs goes to infinity, so the ratio again tends to 0.—Emil J. 18:42, 10 December 2012 (UTC) Never mind[reply]

Sorry, what I mean is, for all pairs {x, y} for x in range(-h, h) for y in range(-h, h) as h approaches infinity. 149.169.218.138 (talk) 18:43, 10 December 2012 (UTC)[reply]

I just had a look one question above, and it struck me that I probably completely misunderstood your question. Are you in fact not asking about the ratio of x/y to x^y, but a discrete version of AH1’s question 2? Is that a coincidence?—Emil J. 19:06, 10 December 2012 (UTC)[reply]

No, it's not a coincidence. Like StuRat said, the continuous version is probably just 1 or 0, so not terribly interesting. The discrete version however, appears to converge. 149.169.218.138 (talk) 19:17, 10 December 2012 (UTC)[reply]

If true, I would rather take that as an indication that StuRat may be wrong. However, note that even though it was not stated in the question, the continuous version only makes sense when x is restricted to be positive, and it is not clear from the wording whether y is intended to be positive as well or not. This may lead to a different discretization than you suggest.—Emil J. 20:00, 10 December 2012 (UTC)[reply]

December 11

Holomorph

Let H be a group of order n, K = Aut(H), and G = H x K where x is the semidirect product with respect to the identity homomorphism $\varphi$ . Let G act on the set X of left cosets of K in G by left multiplication, inducing a permutation representation π from G into S_n. Why do we have that

\pi (G)=N_{S_{n}}(\pi (H))?

I've shown that the left side is a subset of the right, but I can't get the reverse inclusion. The textbook hints that

C_{K}(H)={\text{ker}}\varphi =\{1\}

C_{H}(K)=N_{H}(K)

could be helpful for proving

|G|=|N_{S_{n}}(\pi (H))|,

but I can't see why. Thanks in advance for any help. —Anonymous Dissident^Talk 13:54, 11 December 2012 (UTC)[reply]

Every prime has a Fibonacci divisible by it?

I'm trying to figure out the following question (The answer may be at Fibonacci number#Prime divisors of Fibonacci numbers, but I'm not seeing it.) Is the following true: For all prime p there exists k such that p|F_k . If so, is there a way to calculate a k so it is true? If so, is there a way to calculate the smallest k so that it is true?

Yes, I think it is true. The section that you linked to says that

F_{p-\left({\frac {p}{5}}\right)}\equiv 0{\pmod {p}}

. In other words, for prime p, one of the five Fibonacci numbers between

F_{p-2}

and

F_{p+2}

inclusive will be divisible by p. This is not necessarily the smallest such Fibonacci number - for example, for p = 17,

F_{18}

is divisible by 17, but so is

F_{9}

. Gandalf61 (talk) 16:21, 11 December 2012 (UTC)[reply]

Slight correction (I think): since

\left({\frac {p}{5}}\right)

= 0, 1, or –1, one of the three Fibonacci numbers between

F_{p-1}

and

F_{p+1}

inclusive will be divisible by p. Duoduoduo (talk) 17:16, 11 December 2012 (UTC)[reply]

Yes. 2 divides F₃ and 5 divides F₅. If p is a prime of form 10n+1 or 10n+9 (meaning the last decimal digit is 1 or 9), then p divides F_p−1. If p is a prime of form 10n+3 or 10n+7 (last digit 3 or 7), then p divides F_p+1. PrimeHunter (talk) 03:39, 12 December 2012 (UTC)[reply]

Reciprocals of primes

How did Euler prove the sum of the reciprocals of the primes diverges? I know he created the Riemann zeta function for it but how did he do this? — Preceding unsigned comment added by 86.185.99.78 (talk) 21:18, 11 December 2012 (UTC) — Preceding unsigned comment added by 86.185.99.78 (talk)

See Divergence of the sum of the reciprocals of the primes. Duoduoduo (talk) 21:35, 11 December 2012 (UTC)[reply]

These proofs are interesting but none address how he developed the Riemann zeta function to help his argument. Some of them use the sum of 1/n but none use 1/(n^s) — Preceding unsigned comment added by 86.185.99.78 (talk) 23:19, 11 December 2012 (UTC)[reply]

I believe he got interested in it because of the Basel problem. Dmcq (talk) 01:54, 12 December 2012 (UTC)[reply]

December 12

December 13

Bailey Welch rule.

What is the bailey-welch rule. It is referred to in most papers and research notes related to patients appointment scheduling, but I can't find anywhere on net what this rule actually is. As a guide, it has to do with optimization, operations research,stochastic modelling, possibly a bit of queuing theory etc.

@@ Line 82: / Line 82: @@
 = December 13 =
+== Bailey Welch rule. ==
+What is the bailey-welch rule. It is referred to in most papers and research notes related to patients appointment scheduling, but I can't find anywhere on net what this rule actually is. As a guide, it has to do with optimization, operations research,stochastic modelling, possibly a bit of queuing theory etc.